Question

Given circle with centre O and BC$\parallel$ED, m(arc BC) = 94$°$,m(arc AD) = 86$°$ $\angle$ADE = 8$°$
Find (i) m(arc AE)
(ii) m(arc DC)
(iii) m(arc EB)
Also find $\angle$DAB,$\angle$ECB,$\angle$CBE

Answer 1

Construction: Join OC, OB, OD, OE, OA and PW.

m(arc BC ) = 94$°$
m(arc AD) = 86$°$
$\angle$ADE = 8$°$

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{the}\mathrm{arc}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{at}\mathrm{any}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \left(\mathrm{i}\right) \\ \mathrm{m}(\mathrm{arc}\mathrm{AE})=2\angle \mathrm{ADE}=2\times 8°=16° \\ \left(\mathrm{ii}\right) \\ \angle \mathrm{EOD}=\angle \mathrm{AOD}-\mathrm{AOE}=86°-16°=70° \\ \mathrm{In}△\mathrm{BOC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BOC}+\angle \mathrm{OBC}+\angle \mathrm{OCB}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{OBC}=180°-\angle \mathrm{BOC}(\because \angle \mathrm{OBC}=\angle \mathrm{OCB}) \\ \Rightarrow 2\angle \mathrm{OBC}=180°-94° \\ \Rightarrow \angle \mathrm{OBC}=43° \\ \mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{PQ}\mathrm{passing}\mathrm{through}\mathrm{O}\mathrm{and}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}\mathrm{and}\mathrm{ED}\mathrm{at}\mathrm{P}\mathrm{and}\mathrm{Q},\mathrm{respectively}. \\ \angle \mathrm{POC}=\frac{1}{2}\times 94°=47°(\because ∆\mathrm{BOC}\mathrm{is}\mathrm{an}\mathrm{isoceles}\mathrm{traingle},\angle \mathrm{POC}=\frac{1}{2}\angle \mathrm{BOC}) \\ \angle \mathrm{DOQ}=\frac{1}{2}\times 70°=35°(\because ∆\mathrm{DOE}\mathrm{is}\mathrm{an}\mathrm{isoceles}\mathrm{traingle},\angle \mathrm{DOQ}=\frac{1}{2}\angle \mathrm{BOC}) \\ \angle \mathrm{POC}+\angle \mathrm{COD}+\angle \mathrm{DOQ}=180°(\because \mathrm{POQ}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}) \\ \Rightarrow \angle \mathrm{COD}=180°-\angle \mathrm{POC}-\angle \mathrm{DOQ} \\ \Rightarrow \angle \mathrm{COD}=180°-47°-35°=98° \\ \mathrm{Hence},\mathrm{m}(\mathrm{arc}\mathrm{CD})=98° \\ \left(\mathrm{iii}\right) \\ \angle \mathrm{BOC}+\angle \mathrm{COD}+\angle \mathrm{AOD}+\angle \mathrm{BOA}=360°(\mathrm{Complete}\mathrm{angle}) \\ \Rightarrow \angle \mathrm{BOA}=180°-\angle \mathrm{BOC}-\angle \mathrm{AOD}-\angle \mathrm{COD} \\ \Rightarrow \angle \mathrm{BOA}=180°-94°-86°-98°=82° \\ \mathrm{As}\mathrm{we}\mathrm{know},\mathrm{m}(\mathrm{arc}\mathrm{EB})=\mathrm{m}(\mathrm{arc}\mathrm{AE})+\mathrm{m}(\mathrm{arc}\mathrm{AB})=16°+82°=98° \end{gathered}$$

$$\begin{gathered} \mathrm{Also},\mathrm{m}(\mathrm{arc}\mathrm{DB})=\angle \mathrm{DOC}+\angle \mathrm{COB}=98°+94°=192° \\ \therefore \angle \mathrm{DAB}=\frac{1}{2}\mathrm{m}(\mathrm{arc}\mathrm{DB})=\frac{1}{2}\times 192°=96° \\ \mathrm{m}(\mathrm{arc}\mathrm{EB})=\angle \mathrm{EOA}+\angle \mathrm{AOB}=16°+82°=98° \\ \therefore \angle \mathrm{ECB}=\frac{1}{2}\mathrm{m}(\mathrm{arc}\mathrm{EB})=\frac{1}{2}\times 98°=49° \\ \mathrm{m}(\mathrm{arc}\mathrm{CE})=\angle \mathrm{COD}+\angle \mathrm{DOE}=168° \\ \therefore \angle \mathrm{CBE}=\frac{1}{2}\mathrm{m}(\mathrm{arc}\mathrm{CE})=\frac{1}{2}\times 168°=84° \end{gathered}$$